Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 39

{$(-\dfrac{5}{2},\dfrac{1}{4}),(-4,1)$}

Let us multiply the first equation by $-2$ .After adding the two equations, we get $x=-2(x+3)^2-2$ or, $2x^2+13x+20=0$ or, $(2x+5)(x+4)=0$; or $x=${$-\dfrac{5}{2},-4$} From second equation when $x=-\dfrac{5}{2}$, we have $y=\dfrac{1}{4}$ From second equation when $x=-4$, we have $y=1$ Thus, solution set is: {$(-\dfrac{5}{2},\dfrac{1}{4}),(-4,1)$}